If `H_(2)S` gas is passed through a solution of `K_(2)Cr_(2)O_(7)`, the colour of the solution will -

Options

A. Remain unchanged
B. Become deep red
C. Become green
D. Become deep grey

Answer

To solve the question regarding the color change when `H₂S` gas is passed through a solution of `K₂Cr₂O₇`, we will analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactants in this reaction are:
- `K₂Cr₂O₇` (Potassium dichromate), which is typically orange in color.
- `H₂S` (Hydrogen sulfide) gas.

### Step 2: Understand the Reaction Conditions
The reaction occurs in an acidic medium, often using `H₂SO₄` (sulfuric acid) to provide the necessary acidic environment.

### Step 3: Write the Balanced Chemical Reaction
When `H₂S` is passed through the acidic solution of `K₂Cr₂O₇`, the following reaction occurs:
\[ K_2Cr_2O_7 + H_2S + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + S + H_2O \]
In this reaction:
- Chromium in `K₂Cr₂O₇` is reduced from +6 oxidation state to +3.
- Sulfur in `H₂S` is oxidized from -2 oxidation state to 0 (elemental sulfur).

### Step 4: Analyze the Color Change
- The initial color of the solution is orange due to the presence of `Cr₂O₇²⁻` ions.
- As the chromium is reduced from +6 to +3, the color changes to green due to the formation of `Cr³⁺` ions.

### Step 5: Conclusion
Thus, when `H₂S` gas is passed through a solution of `K₂Cr₂O₇`, the color of the solution changes from orange to green.

### Final Answer
The color of the solution will change from orange to green.

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